(t+h)(x+2)
3h-5h + 11 = 17 is ------2h + 11 = 17- 11 -11____________-2h = 6___ ___-2 -2h = -3 (This is the answer.)
- 3h - 2h + 6h + 9 = h + 9
8h - 10h = 3h + 25-2h = 3h + 25-5h = 25-h = 5h = -5
When h=10, 2h-9 = 2(10)-9 = 20-9 = 11 .
2h + 2h + 2h = 6h
6h 8min
(t+h)(x+2)
3h-5h + 11 = 17 is ------2h + 11 = 17- 11 -11____________-2h = 6___ ___-2 -2h = -3 (This is the answer.)
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
2Si +2HF-2siF + 2H
The complete ionic equation for this reaction is: 2H⁺(aq) + SO₄²⁻(aq) + Ca²⁺(aq) + 2I⁻(aq) → CaSO₄(s) + 2H⁺(aq) + 2I⁻(aq)
No because it equals zero.
Yes, provided p = perimeter, b= breadth and h = height.
- 3h - 2h + 6h + 9 = h + 9
First set up the equation: 2h+10=40 Then get the variable alone by subtracting 10 from each side: 2h=30 And continue by dividing both sides by two 2h=30 ______ 2 2 h=15 So your answer is h=15
3h + (2h 20m) + 35m = (3+2)h + (20+35)m = 5h + 55m